Starter
A sandwich consists of two slices of bread, 3
slices of meat, and one slice of cheese.
For each of the following amounts,
determine the number of sandwiches that
can be made and what is left over:
 6 bread, 10 meat, 4 cheese slices
 10 bread, 6 meat, 8 cheese slices
 25 bread, 40 meat, 12 cheese slices
Starter

6 bread, 10 meat, 4 cheese slices



10 bread, 6 meat, 8 cheese slices



3 sandwiches
0 bread, 1 meat, 1 cheese
2 sandwiches
6 bread, 0 meat, 6 cheese
25 bread, 40 meat, 12 cheese slices


12 sandwiches
1 bread, 4 meat, 0 cheese
Ch. 9 Stoichiometry
9.3 Limiting Reactant
Why is there a limiting reactant?
a reaction rarely has exactly the right
amount of each reactant
 usually have some left over
 limiting reactant

reactant that limits the amount of product
created
 always completely used up


excess reactant

reactant not completely used up
When do you have to find a LR?

whenever two amounts of reactants are
given in a problem

when only one amount of reactant is given
in a problem, then the other is assumed to
be in excess
Finding Limiting Reactant using
reactants
Figure out how much you need of B if you use
up all of A
1.


Convert grams A to grams B using stoichiometry
You may start with either reactant
Determine whether you will have enough
2.


If you don’t have enough of B, then B is LR
If you don’t have enough of A, then A is LR
Finding Limiting Reactant using
products
Convert each of the reactant amounts into
an amount of product (doesn’t matter
which product)
 Compare product amounts and find lowest
amount.
 Whichever reactant led to lowest product
amount is LR

Example 1
The reaction begins with 2.51 g of HF and
4.56 g of SiO2. What is the limiting
reactant and the excess reactant? How
much excess reactant will be left over?
 Write the balanced chemical equation
SiO2(s) + 4HF(g)  SiF4(g) + 2H2O(l)

Example 1

Find the number of moles available of each
reactant:
1 mol HF
2.51 g HF 
 0.125 mol HF
(1.00794  18.99840)g HF
1 mol SiO 2
4.56 g SiO 2 
 0.0759 mol SiO 2
(28.0855  2 *15.9994)g SiO 2
Example 1

If we use up all of the HF, how much SiO2 will we
need to go with it?
1 mol SiO 2
0.125 mol HF 
 0.0313 mol SiO 2 needed
4 mol HF

Do we have enough SiO2?
 0.0759 mol available > 0.0313 mol needed
 YES- there will be some left over
 Limiting Reactant : HF
How much of the product can be
formed?
1.
2.
Start conversion with amount of limiting
reactant.
Convert to amount of product using
stoichiometry
Example 1




How many grams of water could be formed?
Convert grams of HF to moles.
Convert moles of HF to moles of water.
Convert moles to grams using molar mass.
Example 2
A reaction was done with 36.8 g C6H6 and
41.0 g of O2.
 Write the balanced chemical equation
2C6H6 + 15O2  12CO2 + 6H2O
 What is the limiting reactant and how
much of each product can be produced?

Example 2
Reactant Method
36.8 g C6H6
1 mol C6H6
15 mol O2
32 g O2
78.12 g C6H6
2 mol C6H6
1 mol O2
= 113 g O2
Because 113 g O2 is greater than what we have available, O2 is LR
Product Method
36.8 g C6H6
41.0 g O2
1 mol C6H6
6 mol H2O
18.02 g H2O
78.12 g C6H6
2 mol C6H6
1 mol H2O
1 mol O2
6 mol H2O
18.02 g H2O
32 g O2
15 mol O2
1 mol H2O
Because 9.24 is less than 25.5, O2 is the LR
= 25.5 g H2O
= 9.24 g H2O
Example 2

Once you find the LR, you can then calculate
the amount of each product formed…
Amount of H2O
41.0 g O2
1 mol O2
6 mol H2O
18.02 g H2O
32 g O2
15 mol O2
1 mol H2O
1 mol O2
12 mol CO2
44.01 g CO2
32 g O2
15 mol O2
1 mol CO2
= 9.24 g H2O
Amount of CO2
41.0 g O2
= 45.11 g CO2
Example 3

If the reaction below begins with 51.03 grams of Fe
and 37.5 grams of oxygen, what is the limiting
reactant?
4Fe(s) + 3O2(g)  2Fe2O3
Example 3

How many grams of oxygen will be left over
after the reaction?

How many grams of iron (III) oxide can be
formed?